e a s a f t dt f t dt = p = p. t = a

Transcription

1 Mah Fri Apr 7 5, EP76 Today we finish discussing Laplace ransform echniques: Impulse forcing ("dela funcions")oday's noes Convoluion formulas o solve any inhomogeneous consan coefficien linear DE, wih applicaions o ineresing forced oscillaion problemsoday's noes Laplace able enries for oday: f wih f Ce M F s f e s d for s M commens u a uni sep funcion e a s s for urning componens on and off a = a f a u a e a s F s more complicaed on/off a e a s uni impulse/dela "funcion" f g d F s G s convoluion inegrals o inver Laplace ransform producs EP 76 impulse funcions and he operaor Consider a force f acing on an objec for only on a very shor ime inerval a a, for example as when a ba his a ball This impulse p of he force is defined o be he inegral p a a f d and i measures he ne change in momenum of he objec since by Newon's second law m v = f a a m v d = m v a a a = a f d = p Since he impulse p only depends on he inegral of f, and since he exac form of f is unlikely o be known in any case, he easies model is o replace f wih a consan force having he same oal impulse, ie o se f = p d a, where d a, is he uni impulse funcion given by = p

2 d a, =, a, a a, a Noice ha a d a, d = a d = a a Here's a graph of d 2,, for example: Since he uni impulse funcion is a linear combinaion of uni sep funcions, we could solve differenial equaions wih impulse funcions so-consruced As far as Laplace ransform goes, i's even easier o ake he limi as for he Laplace ransforms d a, s, and his effecively models impulses on very shor ime scales d a, = u a u a d a, s = e a s s s e a s = e a s e s In Laplace land we can use L'Hopial's rule (in he variable ) o ake he limi as : lim e a s e s s = e a s s e lim s s s = e a s The resul in ime space is no really a funcion bu we call i he "dela funcion" a anyways, and visualize i as a funcion ha is zero everywhere excep a = a, and ha i is infinie a = a in such a way ha is inegral over any open inerval conaining a equals one As explained in EP76, he dela "funcion" can be hough of in a rigorous way as a linear ransformaion, no as a funcion I can also be hough of as he derivaive of he uni sep funcion u a, and his is consisen wih he Laplace able enries for derivaives of funcions In any case, his leads o he very useful Laplace ransform able enry a uni impulse funcion e a s for impulse forcing

3 Exercise ) Revisi he swing from Wednesday's noes and solve he IVP below for x In his case he paren is providing an impulse each ime he child passes hrough equilibrium posiion afer compleing a cycle x x = x = x = impulse soluion: five equal impulses o ge same final ampliude of meers - Exercise : f 2 Pi sum Heaviside k 2 Pi sin k 2 Pi, k = 4 : plo f, = 2 Pi, color = black, ile = `lazy paren on Friday` ; wih plos : plo plo sin, = Pi, color = black : plo2 plo Pi sin, = Pi 2 Pi, color = black : plo3 plo Pi, = Pi 2 Pi, color = black, linesyle = 2 : plo4 plo Pi, = Pi 2 Pi, color = black, linesyle = 2 : plo5 plo, = Pi, color = black, linesyle = 2 : plo6 plo, = Pi, color = black, linesyle = 2 : display plo, plo2, plo3, plo4, plo5, plo6, ile = `Wednesday advenures a he swingse` ; Wednesday advenures a he swingse lazy paren on Friday

4 Or, an impulse a = and anoher one a = g 2 Pi 2 sin 3 Heaviside Pi sin Pi : plo g, = 2 Pi, color = black, ile = `very lazy paren` ; 3 3 very lazy paren Convoluions and soluions o non-homogeneous physical oscillaion problems (EP76 p 499-5) Consider a mechanical or elecrical forced oscillaion problem for x, and he paricular soluion ha begins a res: a x b x c x = f x = x = Then in Laplace land, his equaion is equivalen o a s 2 X s b s X s c X s = F s X s a s 2 b s c = F s X s = F s a s 2 F s W s b s c Because of he convoluion able enry f g d F s G s convoluion inegrals o inver Laplace ransform producs he soluion is given by x = f w = w f = w f d where w = W s The funcion w is called he "weigh funcion" of he differenial equaion, because he soluion x is some sor of weighed average of he he forces f beween imes and, where he weighing facors are given by w in some sor of convolued way This idea generalizes o much more complicaed mechanical and circui sysems, and is how engineers experimen mahemaically wih how proposed configuraions will respond o various inpu forcing funcions, once hey figure ou he weigh funcion for heir sysem The mahemaical jusificaion for he general convoluion able enry is a he end of oday's noes, for hose who have sudied ieraed double inegrals and who wish o undersand i

5 Exercise 2 Le's play he resonance game and pracice convoluion inegrals, firs wih an old friend, bu hen wih non-sinusoidal forcing funcions We'll sick wih our earlier swing, bu consider various forcing periodic funcions f x x = f x = x = a) Find he weigh funcion w b) Wrie down he soluion formula for x as a convoluion inegral c) Work ou he special case of X s when f = cos, and verify ha he convoluion formula reproduces he answer we would've goen from he able enry 2 k sin k s s 2 k 2 2 sin cos d ; cos sin d ; #convoluion is commuaive 2 sin 2 sin d) Then play he resonance game on he following pages wih new periodic forcing funcions ()

6 We worked ou ha he soluion o our DE IVP will be Example ) A square wave forcing funcion wih ampliude and period 2 Le's alk abou how we came up wih he formula (which works unil = ) wih plos : x = sin f d Since he unforced sysem has a naural angular frequency =, we expec resonance when he forcing funcion has he corresponding period of T = 2 = 2 We will discover ha here is he possibiliy w for resonance if he period of f is a muliple of T (Also, forcing a he naural period doesn' guaranee resonancei depends wha funcion you force wih) f 2 n = n Heaviside n Pi : ploa plo f, = 3, color = green : display ploa, ile = `square wave forcing a naural period` ; square wave forcing a naural period 2 3 ) Wha's your voe? Is his square wave going o induce resonance, ie a response wih linearly growing ampliude? x sin f d : plob plo x, = 3, color = black : display ploa, plob, ile = `resonance response?` ; resonance response? 2 3

7 Example 2) A riangle wave forcing funcion, same period f2 f s ds 5 : # his aniderivaive of square wave should be riangle wave plo2a plo f2, = 3, color = green : display plo2a, ile = `riangle wave forcing a naural period` ; 2) Resonance? riangle wave forcing a naural period 2 3 x2 sin f2 d : plo2b plo x2, = 3, color = black : display plo2a, plo2b, ile = `resonance response?` ; resonance response? 2 3

8 Example 3) Forcing no a he naural period, eg wih a square wave having period T = 2 f3 2 2 n = n Heaviside n : plo3a plo f3, = 2, color = green : display plo3a, ile = `periodic forcing, no a he naural period` ; 3) Resonance? periodic forcing, no a he naural period x3 sin f3 d : plo3b plo x3, = 2, color = black : display plo3a, plo3b, ile = `resonance response?` ; resonance response? 5 5 2

9 Example 4) Forcing no a he naural period, eg wih a paricular wave having period T = 6 f4 n = Heaviside 6 n Heaviside 6 n : plo4a plo f4, = 5, color = green : display plo4a, ile = sporadic square wave wih period 6 ; sporadic square wave wih period ) Resonance? x4 sin f4 d : plo4b plo x4, = 5, color = black : display plo4a, plo4b, ile = `resonance response?` ; resonance response? 5 5

10 Hey, wha happened???? How do we need o modify our hinking if we force a sysem wih somehing which is no sinusoidal, in erms of worrying abou resonance? In he case ha his was modeling a swing (pendulum), how is i geing pushed? Precise Answer: I urns ou ha any periodic funcion wih period P is a (possibly infinie) superposiion P of a consan funcion wih cosine and sine funcions of periods P, 2, P 3, P, Equivalenly, hese 4 funcions in he superposiion are, cos, sin, cos 2, sin 2, cos 3, sin 3, wih ω = 2 This is he P heory of Fourier series, which you will sudy in oher courses, eg Mah 35, Parial Differenial Equaions If he given periodic forcing funcion f has non-zero erms in his superposiion for which n = (he naural angular frequency) (equivalenly P n = 2 = T ), here will be resonance; oherwise, no resonance We could already have undersood some of his in Chaper 5, for example Exercise 3) The naural period of he following DE is (sill) T = 2 forcing funcion below is T = 6 Noice ha he period of he firs and ha he period of he second one is T = T = 2 Ye, i is he firs DE whose soluions will exhibi resonance, no he second one Explain, using Chaper 5 superposiion ideas a) x x = cos sin 3 b) x x = cos 2 3 sin 3

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12 Mah Week 3 Mon Apr : We will use las Friday's noes o discuss of finish discussing impulse forcing and he convoluion able enry for Laplace ransforms If we have ime, we'll move ahead ino Tuesday's noes Because i is so imporan i's worh highlighing ha for any non-homogeneous forced oscillaion problem, he soluion funcion x for he IVP ha sars wih zero iniial condiions can be wrien as a convoluion inegral: For he IVP a x b x c x = f x = x = Then in Laplace land, his equaion is equivalen o a s 2 X s b s X s c X s = F s X s a s 2 b s c = F s X s = F s a s 2 b s c = F s W s for W s a s 2 b s c The inverse Laplace ransform w = W s is called he "weigh funcion" of he given differenial equaion Noice (check!) ha w is he soluion o he homogeneous DE IVP a x b x c x = x = x = Because of he convoluion able enry f g d F s G s convoluion inegrals o inver Laplace ransform producs he soluion (for ANY forcing funcion f ) is given by x = f w d (Wih non-zero iniial condiions here would be homogeneous soluion erms as well In he case of damping hese erms would be ransien) Noice ha his says ha x depends on he values of he forcing funcion f for he previous imes, weighed by w, Tha he non-homogenous soluions can be consruced from he homogeneous ones via his convoluion is a special case of "Duhamel's Principle", which applies o linear DE's and linear PDE's: hps://enwikipediaorg/wiki/duhamel%27s_principle

13 Mah Tues Apr 6-62 Eigenvalues and eigenvecors for square marices The sudy of eigenvalues and eigenvecors is a reurn o marix linear algebra, and he conceps we discuss will help us sudy linear sysems of differenial equaions, in Chaper 7 Such sysems of DE's arise naurally in he conexs of coupled inpu-oupu models, wih several componens coupled mass-spring or RLC circui loops, wih several componens To inroduce he idea of eigenvalues and eigenvecors we'll firs hink geomerically Example Consider he marix ransformaion T : 2 2 wih formula T x = 3 x 3 = x x 2 Noice ha for he sandard basis vecors e =, T, e 2 =, T T e = 3e T e 2 = e 2 The facs ha T is linear and ha i ransforms e, e 2 by scalar muliplying hem, les us undersand he geomery of his ransformaion compleely: x T = T x e e 2 = x T e T e 2 = x 3e e 2 In oher words, T sreches by a facor of 3 in he e direcion, and by a facor of in he e 2 direcion, ransforming a square grid in he domain ino a parallel recangular grid in he image:

14 Exercise ) Do a similar geomeric analysis and skech for he ransformaion T x = 2 3 x Exercise 2) And for he ransformaion T x = 2 x Definiion: If A n n and if A v = v for a scalar and a vecor v hen v is called an eigenvecor of A, and is called he eigenvalue of v (In some exs he words characerisic vecor and characerisic value are used as synonyms for hese words) In he hree examples above, he sandard basis vecors (or muliples of hem) were eigenvecors, and he corresponding eigenvalues were he diagonal marix enries A non-diagonal marix may sill have eigenvecors and eigenvalues, and his geomeric informaion can sill be imporan o find Bu how do you find eigenvecors and eigenvalues for non-diagonal marices?

15 Exercise 3) Try o find eigenvecors and eigenvalues for he non-diagonal marix, by jus rying random inpu vecors x and compuing A x A = How o find eigenvalues and eigenvecors (including eigenspaces) sysemaically: If A v = v A v v = A v I v = where I is he ideniy marix A I v = As we know, his las equaion can have non-zero soluions v if and only if he marix A I is no inverible, ie de A I = So, o find he eigenvalues and eigenvecors of marix you can proceed as follows: Compue he polynomial in λ p = de A I If A n n hen p will be degree n This polynomial is called he characerisic polynomial of he marix A j can be an eigenvalue for some non-zero eigenvecor v if and only if i's a roo of he characerisic polynomial, ie p will be eigenvecors wih eigenvalue since A j j = For each such roo, he homogeneous soluion space of vecors v solving A j I v = This subspace of eigenvecors will be a leas one dimensional, j I does no reduce o he ideniy and so he explici homogeneous soluions will have free parameers Find a basis of eigenvecors for his subspace Follow his procedure for each eigenvalue, ie for each roo of he characerisic polynomial Noaion: The subspace of eigenvecors for eigenvalue j is called he j eigenspace, and denoed by E j (We include he zero vecor in E ) The basis of eigenvecors is called an eigenbasis for E j j

16 Exercise 4) a) Use he sysemaic algorihm o find he eigenvalues and eigenbases for he non-diagonal marix of Exercise 3 A = b) Use your work o describe he geomery of he linear ransformaion in erms of direcions ha ge sreched: T x = x Exercise 5) Find he eigenvalues and eigenspace bases for 4 2 B := (i) Find he characerisic polynomial and facor i o find he eigenvalues (ii) for each eigenvalue, find bases for he corresponding eigenspaces (iii) Can you describe he ransformaion T x = Bx geomerically using he eigenbases? Does de B have anyhing o do wih he geomery of his ransformaion?

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18 Your soluion will be relaed o he oupu below: In all of our examples so far, i urns ou ha by collecing bases from each eigenspace for he marix A n n, and puing hem ogeher, we ge a basis for n This les us undersand he geomery of he ransformaion T x = A x almos as well as if A is a diagonal marix This is acually somehing ha does no always happen for a marix A When i does happen, we say ha A is diagonalizable Here's an example of a marix which is NOT diagonalizable: Exercise 7: Find marix eigenvalues and eigenspace basis for each eigenvalue, for A = Explain why here is no basis of 2 consising of eigenvecors of A

19 Mah Wed Apr Eigenvalues and eigenvecors for square marices; diagonalizabiliy of marices Recall from yeserday, Definiion: If A n n and if A v = v for some scalar and vecor v hen v is called an eigenvecor of A, and is called he eigenvalue of v (and an eigenvalue of A) For general marices, he eigenvecor equaion A v = v can be rewrien as A I v = The only way such an equaion can hold for v is if he marix A I does no reduce o he ideniy marix In oher words - de A I mus equal zero Thus he only possible eigenvalues associaed o a given marix mus be roos of he characerisic polynomial j p = de A I So, he firs sep in finding eigenvecors for A is acually o find he eigenvalues - by finding he characerisic polynomial and is roos j For each roo j he marix A j I will no reduce o he ideniy, and he soluion space o A j I v = will be a leas one-dimensional, and have a basis of one or more eigenvecors Find such a basis for his j eigenspace E by reducing he homogeneous marix equaion = j A j I v =, backsolving, and exracing a basis We can ofen "see" and eigenvecor by realizing ha homogeneous soluions o a marix equaion correspond o column dependencies Finish any lefover exercises from Tuesday

20 Exercise ) If your marix A is diagonal, he general algorihm for finding eigenspace bases jus reproduces he enries along he diagonal as eigenvalues, and he corresponding sandard basis vecors as eigenspace bases (Recall our diagonal marix examples from yeserday, where he sandard basis vecors were eigenvecors This is ypical for diagonal marices) Illusrae how his works for a 3 3 diagonal marix, so ha in he fuure you can jus read of he eigendaa if he marix you're given is (already) diagonal: a A a 22 a 33 sep ) Find he roos of he characerisic polynomial de A I sep 2) Find he eigenspace bases, assuming he values of a, a 22, a 33 are disinc (all differen) Wha if a = a 22 bu hese values do no equal a 33?

21 In all of our examples so far, i urns ou ha by collecing bases from each eigenspace for he marix A n n, and puing hem ogeher, we ge a basis for n This les us undersand he geomery of he ransformaion T x = A x almos as well as if A is a diagonal marix, and so we call such marices diagonalizable Having such a basis of eigenvecors for a given marix is also exremely useful for algebraic compuaions, and will give anoher reason for he word diagonalizable o describe such marices Use he 3 basis made of ou eigenvecors of he marix B in Exercise, and pu hem ino he columns of a marix we will call P We could order he eigenvecors however we wan, bu we'll pu he E = 2 basis vecors in he firs wo columns, and he E = 3 basis vecor in he hird column: P := 2 2 Now do algebra (check hese seps and discuss wha's going on!) = = In oher words, B P = P D, where D is he diagonal marix of eigenvalues (for he corresponding columns of eigenvecors in P) Equivalenly (muliply on he righ by P or on he lef by P ): B = P D P and P BP = D Exercise 2) Use one of he he ideniies above o show how B can be compued wih only wo marix muliplicaions!

22 Definiion: Le A n n If here is an n (or n basis v, v 2,, v n consising of eigenvecors of A, hen A is called diagonalizable This is precisely why: Wrie A v j = j v j (some of hese j may be he same, as in he previous example) Le P be he marix P = v v 2 v n Then, using he various ways of undersanding marix muliplicaion, we see A P = A v v 2 v n = = v v 2 v n v 2 v 2 n v n 2 : : : A P = P D A = P D P P A P = D n Unforunaely, no all marices are diagonalizable: Exercise 3) Show ha is no diagonalizable C := 2 2 3

23 Facs abou diagonalizabiliy (see ex secion 62 for complee discussion, wih reasoning): Le A n n have facored characerisic polynomial k k 2 k m p = n where like erms have been colleced so ha each because he degree of p is n j 2 is disinc (ie differen) Noice ha k k 2 k m = n Then dim E = j k j If dim E = j k j hen he j m eigenspace is called defecive The marix A is diagonalizable if and only if each dim E = j = k j In his case, one obains an n eigenbasis simply by combining bases for each eigenspace ino one collecion of n vecors (Laer on, he same definiions and reasoning will apply o complex eigenvalues and eigenvecors, and a basis of n ) In he special case ha A has n disinc eigenvalues, 2,, each eigenspace is forced o be n dimensional since k k 2 k n = n so each k j = Thus A is auomaically diagonalizable as a special case of he second bulle poin Exercise 4) How do he examples from oday and yeserday compare wih he general facs abou diagonalizabiliy?

24 Mah Fri Apr 4 7 Sysems of differenial equaions - o model muli-componen sysems via comparmenal analysis: hp://enwikipediaorg/wiki/muli-comparmen_model Here's a relaively simple 2-ank problem o illusrae he ideas: Exercise ) Find differenial equaions for solue amouns x, above, using inpu-oupu modeling Assume solue concenraion is uniform in each ank If x = b, = b 2, wrie down he iniial value problem ha you expec would have a unique soluion answer (in marix-vecor form): 4 2 x = 4 2 x x = b b 2

25 Geomeric inerpreaion of firs order sysems of differenial equaions The example on page is a special case of he general iniial value problem for a firs order sysem of differenial equaions: x = F, x x = x We will see how any single differenial equaion (of any order), or any sysem of differenial equaions (of any order) is equivalen o a larger firs order sysem of differenial equaions And we will discuss how he naural iniial value problems correspond Why we expec IVP's for firs order sysems of DE's o have unique soluions x : From eiher a mulivariable calculus course, or from physics, recall he geomeric/physical inerpreaion of x as he angen/velociy vecor o he parameric curve of poins wih posiion vecor x, as varies This picure should remind you of he discussion, bu ask quesions if his is new o you: Analyically, he reason ha he vecor of derivaives x compued componen by componen is acually a limi of scaled secan vecors (and herefore a angen/velociy vecor) is: x x x lim : : x n x n = lim x x : x n x n provided each componen funcion is differeniable Therefore, he reason you expec a unique soluion o he IVP for a firs order sysem is ha you know where you sar (x = x ), and you know your "velociy" vecor (depending on ime and curren locaion) you expec a unique soluion! (Plus, you could use somehing like a vecor version of Euler's mehod or he Runge-Kua mehod o approximae i! You jus conver he scalar quaniies in he code ino vecor quaniies And his is wha numerical solvers do) = x x n :,

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27 Exercise 2) Reurn o he page ank example x = 4 x 2 = 4 x 2 x = 9 = 2a) Inerpre he parameric soluion curve x, T o his IVP, as indicaed in he pplane screen sho below ("pplane" is he siser program o "dfield", ha we were using in Chapers -2) Noice how i follows he "velociy" vecor field (which is ime-independen in his example), and how he "paricle T moion" locaion x, is acually he vecor of solue amouns in each ank, a ime If your sysem involved en coupled anks raher han wo, hen his "paricle" is moving around in 2b) Wha are he apparen limiing solue amouns in each ank? 2c) How could your smar-alec younger sibling have old you he answer o 2b wihou considering any differenial equaions or "velociy vecor fields" a all?

28 Firs order sysems of differenial equaions of he form x = A x are called linear homogeneous sysems of DE's (Think of rewriing he sysem as x A x = in analogy wih how we wroe linear scalar differenial equaions) Then he inhomogeneous sysem of firs order DE's would be wrien as x A x = f or x = A x f Noice ha he operaor on vecor-valued funcions x defined by L x x A x is linear, ie L x y = L x L y L c x = c L x SO! The space of soluions o he homogeneous firs order sysem of differenial equaions x A x = is a subspace AND he general soluion o he inhomogeneous sysem x A x = f will be of he form x = x P x H where x P is any single paricular soluion and x H is he general homogeneous soluion Exercise 3) In he case ha A is a consan marix (ie enries don' depend on ), consider he homogeneous problem x = A x Look for soluions of he form x = e v, where v is a consan vecor Show ha x = e v solves he homogeneous DE sysem if and only if v is an eigenvecor of A, wih eigenvalue, ie A v = v Hin: In order for such an x o solve he DE i mus be rue ha and Se hese wo expressions equal x = e v A x = A e v = e A v

29 Exercise 4) Use he idea of Exercise 3 o solve he iniial value problem of Exercise 2!! Compare your soluion x o he parameric curve drawn by pplane, ha we looked a a couple of pages back Exercise 5) Lessons learned from ank example: Wha condiion on he marix A n uniquely solve every iniial value problem x = A x x = x n n will allow you o using he mehod in Exercise 3-4? Hin: Chaper 6 (If ha condiion fails here are oher ways o find he unique soluions)

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